Find all functions $f$ such that $f\left(x^2-y^2\right)=(x-y)\big(f(x)+f(y)\big)$.

Let $P(x, y)$ denote $$f(x^2-y^2)=(x-y)(f(x)+f(y))$$

$P(0, 0)$ gives $f(0)=0$, then $P(x, 0)$ gives $f(x^2)=xf(x)$.

Now $-xf(-x)=f((-x)^2)=f(x^2)=xf(x)$. Thus for $x \not =0$, $f(-x)=-f(x)$.

This clearly holds for $x=0$ as well, so $f(-x)=-f(x)$.

$P(x, -y)$ gives $$f(x^2-y^2)=(x+y)(f(x)+f(-y))=(x+y)(f(x)-f(y))$$

Comparing with $P(x, y)$ gives $$(x+y)(f(x)-f(y))=(x-y)(f(x)+f(y))$$

Thus $2yf(x)=2xf(y)$. Putting $y=1$ gives $f(x)=xf(1)$. We may easily verify that these are indeed solutions.

I don't know how to do this without continuity. If you can show $f(x) \to f(0)$ whenever $x \to 0$, then the original expression yields it. Then all you need is $f(x)$ bounded near zero in order to use $f(x^2) = x f(x)$ to get that the limit exists and is zero.

However, taking continuity for granted, consider your result $f(x^2)=xf(x)$ which can be rewritten as $f(x) = x^{1/2} f(x^{1/2})$ if we let $x^2 \to x$ for positive $x$. Then $f(x^2) = x \cdot x^{1/2} f(x^{1/2})$. But then we can write $f(x^{1/2}) = x^{1/4} f(x^{1/4})$ if we let $x \to x^{1/4}$ in $f(x^2) = xf(x)$ so that $f(x^2) = x \cdot x^{1/2} \cdot x^{1/4} f(x^{1/4})$.

Repeating this $n$ times we get $$f(x^2) = x^{\sum_{n=0} 2^{-n}} f(x^{2^{-n}}) = x^{(2-2^{-n})} f(x^{2^{-n}})$$ which holds for $x>0$. Now for $x > 0$, taking the limit as $n \to \infty$ gives $$f(x^2) = x^2 f(1)$$ using continuity of $f$ or restated $$f(x) = f(1) x$$ for positive $x$.

But now for negative $x$ use the fact $f(x) = -f(-x)$ so that $f(-x) = -x f(1)$. Letting $\alpha = f(1)$, we say all solutions must be of the form $f(x) = \alpha x$. We easily verify they are all solutions.

$f(x)+f(x+1)=f(2x+1)=(2x+1)(f(x+1)+f(-x))$.

That can be rearranged to

$$\frac{f(x)}{x}=\frac{f(x+1)}{x+1}$$

so, for example, $f(n)=nf(1)$